WavesClass 11 Physics NCERT Solutions

Given: Mass of the string, $m = 2.50 \text{ kg}$ Tension in the string, $T = 200 \text{ N}$ Length of the string, $L = 20.0 \text{ m}$

To Find: The time taken for the disturbance to reach the other end, $t$.

Formula: The speed of a transverse wave on a string is given by: $v = \sqrt{\frac{T}{\mu}}$ where $\mu$ is the linear mass density, $\mu = \frac{m}{L}$. The time taken is given by: $t = \frac{L}{v}$

Calculation: First, calculate the linear mass density ($\mu$): $\mu = \frac{m}{L} = \frac{2.50 \text{ kg}}{20.0 \text{ m}} = 0.125 \text{ kg/m}$ Next, calculate the speed of the wave ($v$): $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{200 \text{ N}}{0.125 \text{ kg/m}}} = \sqrt{1600} = 40 \text{ m/s}$ Finally, calculate the time taken ($t$): $t = \frac{L}{v} = \frac{20.0 \text{ m}}{40 \text{ m/s}} = 0.5 \text{ s}$

Final Answer: The disturbance takes $0.5 \text{ s}$ to reach the other end.

Given: Height of the tower, $h = 300 \text{ m}$ Speed of sound in air, $v_{sound} = 340 \text{ m/s}$ Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$ Initial velocity of the stone, $u = 0 \text{ m/s}$

To Find: The total time ($t_{total}$) after which the splash is heard at the top.

Analysis: The total time is the sum of two parts:

Formula: For the falling stone, using the equation of motion: $s = ut + \frac{1}{2}at^2$ For the sound travelling up: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

Final Answer: The splash is heard at the top after approximately $8.7 \text{ s}$.

Given: Length of the steel wire, $L = 12.0 \text{ m}$ Mass of the wire, $m = 2.10 \text{ kg}$ Speed of the transverse wave, $v = 343 \text{ m/s}$ (equal to the speed of sound in air at $20^{\circ} \text{C}$)

To Find: The tension in the wire, $T$.

Formula: The speed of a transverse wave on a stretched wire is given by: $v = \sqrt{\frac{T}{\mu}}$ where $\mu$ is the linear mass density, $\mu = \frac{m}{L}$. Rearranging for tension, $T$: $T = v^2 \mu$

Calculation: First, calculate the linear mass density ($\mu$): $\mu = \frac{m}{L} = \frac{2.10 \text{ kg}}{12.0 \text{ m}} = 0.175 \text{ kg/m}$ Now, calculate the tension ($T$): $T = v^2 \mu = (343 \text{ m/s})^2 \times 0.175 \text{ kg/m}$ $T = 117649 \times 0.175 \text{ N}$ $T = 20588.575 \text{ N}$ $T \approx 2.06 \times 10^4 \text{ N}$

Final Answer: The tension in the wire should be approximately $2.06 \times 10^4 \text{ N}$.

The formula for the speed of sound in a gas is $v=\sqrt{\frac{\gamma P}{\rho}}$, where $\gamma$ is the ratio of specific heats, $P$ is the pressure, and $\rho$ is the density of the gas.

(a) Independence of pressure: For an ideal gas, the equation of state is $PV = nRT$. If $M$ is the molar mass of the gas, then the density $\rho$ is given by $\rho = \frac{m}{V} = \frac{nM}{V}$. Substituting $n/V$ from the ideal gas equation, we get: $\rho = \frac{PM}{RT}$ This gives the ratio: $\frac{P}{\rho} = \frac{RT}{M}$ Substituting this into the speed of sound formula: $v = \sqrt{\frac{\gamma RT}{M}}$ This equation shows that for a given gas (constant $\gamma$ and $M$) at a constant temperature ($T$), the speed of sound $v$ is constant. It does not depend on the pressure $P$, because as pressure changes, the density $\rho$ also changes proportionally, keeping the ratio $P/\rho$ constant.

(b) Increase with temperature: From the formula $v = \sqrt{\frac{\gamma RT}{M}}$, it is clear that the speed of sound is directly proportional to the square root of the absolute temperature ($T$ in Kelvin). $v \propto \sqrt{T}$ Therefore, as the temperature of the air increases, the speed of sound in air increases.

(c) Increase with humidity: Humidity refers to the presence of water vapor in the air. The molar mass of water vapor ($M_{H_2O} \approx 18 \text{ g/mol}$) is less than the average molar mass of dry air (mostly nitrogen and oxygen, $M_{air} \approx 29 \text{ g/mol}$). When humidity increases, lighter water molecules replace heavier nitrogen and oxygen molecules. This results in the density of moist air ($\rho_{moist}$) being less than the density of dry air ($\rho_{dry}$) at the same temperature and pressure. From the formula $v = \sqrt{\frac{\gamma P}{\rho}}$, the speed of sound is inversely proportional to the square root of the density. $v \propto \frac{1}{\sqrt{\rho}}$ Since $\rho_{moist} < \rho_{dry}$, the speed of sound in humid air is greater than in dry air.

The converse is true. Any function of the form $y=f(x \pm vt)$ represents a travelling wave. The function describes a shape $y=f(x')$ which propagates along the x-axis without distortion at a speed $v$. However, for a function to represent a physical wave, it must be well-defined and have a finite value for all physically relevant values of $x$ and $t$.

Let's examine the given functions:

(a) $y = (x-v t)^{2}$ This function is of the form $f(x-vt)$, where the function $f(z) = z^2$. This represents a parabolic pulse travelling in the positive x-direction with speed $v$. For all finite values of $x$ and $t$, the displacement $y$ is finite. Thus, it can represent a travelling wave.

(b) $y = \log \left[(x+v t) / x_{0}\right]$ This function is of the form $f(x+vt)$, where the function $f(z) = \log(z/x_0)$. This represents a wave travelling in the negative x-direction with speed $v$. However, the logarithm function is not defined for non-positive arguments and becomes infinite as the argument approaches zero. This means the displacement would be undefined or infinite for certain values of $x$ and $t$ (when $x+vt \le 0$). While it mathematically fits the form, it may not be physically realistic for all conditions, but it does represent a travelling wave.

(c) $y = 1 /(x+v t)$ This function is of the form $f(x+vt)$, where the function $f(z) = 1/z$. This represents a wave travelling in the negative x-direction with speed $v$. This function has a singularity (becomes infinite) at $x+vt=0$. This implies an infinite displacement, which is not physically possible. Despite this physical limitation, it mathematically satisfies the condition for representing a travelling wave.

Conclusion: All three functions can possibly represent a travelling wave because they are functions of the form $f(x \pm vt)$. The physical realizability of these waves depends on whether the displacement remains finite, which is not the case for (b) and (c) under all conditions.

Given: Frequency of the ultrasonic sound, $f = 1000 \text{ kHz} = 1000 \times 10^3 \text{ Hz} = 10^6 \text{ Hz}$ Speed of sound in air, $v_{air} = 340 \text{ m/s}$ Speed of sound in water, $v_{water} = 1486 \text{ m/s}$

Key Concept: When a wave passes from one medium to another, its frequency remains unchanged. The wavelength and speed change.

Formula: The relationship between speed ($v$), frequency ($f$), and wavelength ($\lambda$) is: $v = f \lambda \quad \Rightarrow \quad \lambda = \frac{v}{f}$

(a) Wavelength of the reflected sound: The reflected sound travels back into the air. Therefore, we use the speed of sound in air.

Calculation: $\lambda_{reflected} = \frac{v_{air}}{f} = \frac{340 \text{ m/s}}{10^6 \text{ Hz}}$ $\lambda_{reflected} = 340 \times 10^{-6} \text{ m} = 3.4 \times 10^{-4} \text{ m}$

Final Answer (a): The wavelength of the reflected sound is $3.4 \times 10^{-4} \text{ m}$.

(b) Wavelength of the transmitted sound: The transmitted sound travels into the water. Therefore, we use the speed of sound in water.

Calculation: $\lambda_{transmitted} = \frac{v_{water}}{f} = \frac{1486 \text{ m/s}}{10^6 \text{ Hz}}$ $\lambda_{transmitted} = 1486 \times 10^{-6} \text{ m} = 1.486 \times 10^{-3} \text{ m}$

Final Answer (b): The wavelength of the transmitted sound is $1.486 \times 10^{-3} \text{ m}$.

Given: Speed of sound in the tissue, $v = 1.7 \text{ km/s} = 1.7 \times 10^3 \text{ m/s} = 1700 \text{ m/s}$ Operating frequency of the scanner, $f = 4.2 \text{ MHz} = 4.2 \times 10^6 \text{ Hz}$

To Find: The wavelength of sound in the tissue, $\lambda$.

Formula: The relationship between speed ($v$), frequency ($f$), and wavelength ($\lambda$) is: $v = f \lambda \quad \Rightarrow \quad \lambda = \frac{v}{f}$

Calculation: $\lambda = \frac{1700 \text{ m/s}}{4.2 \times 10^6 \text{ Hz}}$ $\lambda = \frac{170}{4.2} \times 10^{-5} \text{ m}$ $\lambda \approx 40.476 \times 10^{-5} \text{ m}$ $\lambda \approx 4.05 \times 10^{-4} \text{ m}$

Final Answer: The wavelength of sound in the tissue is approximately $4.05 \times 10^{-4} \text{ m}$.

The given equation for the transverse harmonic wave is: $y(x, t) = 3.0 \sin (36t + 0.018x + \pi/4)$ We compare this with the standard form of a wave travelling along the x-axis: $y(x, t) = a \sin (\omega t + kx + \phi)$

(c) Initial Phase at the Origin: The initial phase at the origin ($x=0, t=0$) is the phase constant $\phi$. By comparison, the initial phase is $\phi = \frac{\pi}{4} \text{ rad}$.

(d) Least Distance between Two Successive Crests: The least distance between two successive crests is the wavelength ($\lambda$). The wavelength is related to the angular wave number $k$ by $\lambda = \frac{2\pi}{k}$. $\lambda = \frac{2\pi}{0.018 \text{ rad/cm}} \approx 349 \text{ cm} \approx 3.49 \text{ m}$

Summary of Answers: (a) It is a travelling wave. Speed is $20 \text{ m/s}$. Direction is from right to left (negative x-direction). (b) Amplitude is $3.0 \text{ cm}$. Frequency is approximately $5.73 \text{ Hz}$. (c) Initial phase at the origin is $\frac{\pi}{4} \text{ rad}$. (d) The least distance between two successive crests is approximately $3.49 \text{ m}$.

The equation of the wave is: $y(x, t) = 3.0 \sin (36t + 0.018x + \pi/4)$ where $x, y$ are in cm and $t$ is in s.

To plot the displacement ($y$) versus time ($t$) graphs, we substitute the given values of $x$.

For $x = 0$ cm: $y(0, t) = 3.0 \sin (36t + \pi/4)$ This is a sinusoidal function of time with amplitude 3.0 cm and an initial phase of $\pi/4$.

For $x = 2$ cm: $y(2, t) = 3.0 \sin (36t + 0.018 \times 2 + \pi/4)$ $y(2, t) = 3.0 \sin (36t + 0.036 + \pi/4)$ This is also a sinusoidal function of time with the same amplitude (3.0 cm) and the same angular frequency (36 rad/s). The phase is shifted by an additional $0.036$ radians.

For $x = 4$ cm: $y(4, t) = 3.0 \sin (36t + 0.018 \times 4 + \pi/4)$ $y(4, t) = 3.0 \sin (36t + 0.072 + \pi/4)$ This is again a sinusoidal function of time with the same amplitude and frequency, but with a further phase shift.

Shapes of the graphs: All three graphs are sinusoidal (sine curves). They will have the same amplitude (3.0 cm) and the same period ($T = 2\pi/\omega = 2\pi/36 \approx 0.175$ s). They will be shifted horizontally (in time) with respect to each other due to their different phase constants.

Difference in oscillatory motion: In a travelling wave, the oscillatory motion of different points (different $x$ values) differs in phase.

The given wave equation is: $y(x, t) = 2.0 \cos 2\pi(10t - 0.0080x + 0.35)$ Let us rewrite this in the standard form $y(x, t) = a \cos(\omega t - kx + \phi)$. $y(x, t) = 2.0 \cos(20\pi t - 0.016\pi x + 0.7\pi)$ The phase of the wave is $\Phi(x, t) = 20\pi t - 0.016\pi x + 0.7\pi$.

The phase difference ($\Delta\Phi$) between two points separated by a distance $\Delta x$ at a given time is given by: $\Delta\Phi = |k \Delta x|$ From the equation, the angular wave number $k$ is: $k = 0.016\pi \text{ rad/cm}$

(a) Distance $\Delta x = 4 \text{ m} = 400 \text{ cm}$ $\Delta\Phi = (0.016\pi \text{ rad/cm}) \times (400 \text{ cm})$ $\Delta\Phi = 6.4\pi \text{ rad}$

(b) Distance $\Delta x = 0.5 \text{ m} = 50 \text{ cm}$ $\Delta\Phi = (0.016\pi \text{ rad/cm}) \times (50 \text{ cm})$ $\Delta\Phi = 0.8\pi \text{ rad}$

(c) Distance $\Delta x = \lambda / 2$ The phase difference for a path difference of $\lambda/2$ is always $\pi$ radians. Alternatively, we can calculate it: The wavelength $\lambda$ is related to $k$ by $\lambda = 2\pi/k$. $\Delta\Phi = k \Delta x = k \left(\frac{\lambda}{2}\right) = k \left(\frac{1}{2} \frac{2\pi}{k}\right) = \pi \text{ rad}$

(d) Distance $\Delta x = 3\lambda / 4$ Similarly, for a path difference of $3\lambda/4$: $\Delta\Phi = k \Delta x = k \left(\frac{3\lambda}{4}\right) = k \left(\frac{3}{4} \frac{2\pi}{k}\right) = \frac{3\pi}{2} \text{ rad}$

Final Answers: The phase difference is: (a) $6.4\pi \text{ rad}$ (b) $0.8\pi \text{ rad}$ (c) $\pi \text{ rad}$ (d) $\frac{3\pi}{2} \text{ rad}$

The given equation is $y(x, t) = 0.06 \sin\left(\frac{2\pi}{3}x\right) \cos(120\pi t)$.

(a) Travelling or Stationary Wave? The function represents a stationary wave. This is because the displacement $y$ is a product of two separate functions, one of position $x$ ($0.06 \sin(\frac{2\pi}{3}x)$) and one of time $t$ ($\cos(120\pi t)$). In a travelling wave, $x$ and $t$ appear in the combination $(kx \pm \omega t)$.

(b) Interpretation as Superposition: A stationary wave can be interpreted as the superposition of two identical progressive waves travelling in opposite directions. The standard equation for a stationary wave formed this way is: $y(x, t) = [2a \sin(kx)] \cos(\omega t)$ Comparing this with the given equation: $y(x, t) = [0.06 \sin(\frac{2\pi}{3}x)] \cos(120\pi t)$ We can identify the parameters of the component waves:

Now, we can find the wavelength, frequency, and speed of each component wave:

(c) Tension in the string: Given: Length of the string, $L = 1.5 \text{ m}$ Mass of the string, $m = 3.0 \times 10^{-2} \text{ kg}$

Formula: The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$. Therefore, the tension is $T = v^2 \mu$.

Calculation: First, calculate the linear mass density ($\mu$): $\mu = \frac{m}{L} = \frac{3.0 \times 10^{-2} \text{ kg}}{1.5 \text{ m}} = 2.0 \times 10^{-2} \text{ kg/m}$ Now, calculate the tension ($T$) using the speed $v = 180 \text{ m/s}$ from part (b): $T = (180 \text{ m/s})^2 \times (2.0 \times 10^{-2} \text{ kg/m})$ $T = 32400 \times 2.0 \times 10^{-2} \text{ N}$ $T = 648 \text{ N}$

Final Answer: (a) Stationary wave. (b) Wavelength $\lambda = 3 \text{ m}$, frequency $f = 60 \text{ Hz}$, and speed $v = 180 \text{ m/s}$. (c) The tension in the string is $648 \text{ N}$.

(a) $y = 2 \cos(3x) \sin(10t)$ This function is a product of a function of position $x$ and a function of time $t$. This is the characteristic form of a (ii) stationary wave.

(b) $y = 2 \sqrt{x-vt}$ This function has $x$ and $t$ appearing in the combination $(x-vt)$. It is of the form $f(x-vt)$. This represents a (i) travelling wave moving in the positive x-direction.

(c) $y = 3 \sin(5x - 0.5t) + 4 \cos(5x - 0.5t)$ This is a superposition of two travelling waves, both with the same argument $(5x - 0.5t)$, and therefore the same wavelength and frequency. Such a superposition results in another travelling wave with a different amplitude and phase. We can write this in the form $A \sin(kx - \omega t + \phi)$. Therefore, this function represents a (i) travelling wave.

(d) $y = \cos x \sin t + \cos 2x \sin 2t$ This function is a superposition of two stationary waves, $\cos x \sin t$ and $\cos 2x \sin 2t$. These two stationary waves have different frequencies ($\omega_1 = 1, \omega_2 = 2$) and different wavelengths ($\lambda_1 = 2\pi, \lambda_2 = \pi$). The superposition of two stationary waves with different frequencies does not result in a simple travelling or stationary wave. The positions of nodes and antinodes are not fixed, and the wave pattern does not propagate with a constant shape. Therefore, this represents (iii) none at all (neither a simple travelling nor a stationary wave).

Given: Fundamental frequency, $f_1 = 45 \text{ Hz}$ Mass of the wire, $m = 3.5 \times 10^{-2} \text{ kg}$ Linear mass density, $\mu = 4.0 \times 10^{-2} \text{ kg/m}$

Formula: For a string fixed at both ends, the fundamental frequency is given by: $f_1 = \frac{v}{2L}$ where $v$ is the speed of the wave and $L$ is the length of the string. We can find the length $L$ from the mass and linear mass density: $L = \frac{m}{\mu}$

Calculation: First, calculate the length of the wire ($L$): $L = \frac{3.5 \times 10^{-2} \text{ kg}}{4.0 \times 10^{-2} \text{ kg/m}} = 0.875 \text{ m}$ Now, rearrange the frequency formula to find the speed ($v$): $v = 2 L f_1$ $v = 2 \times 0.875 \text{ m} \times 45 \text{ Hz}$ $v = 1.75 \times 45 \text{ m/s}$ $v = 78.75 \text{ m/s}$

Final Answer (a): The speed of a transverse wave on the string is $78.75 \text{ m/s}$.

Formula: The speed of a transverse wave is also given by: $v = \sqrt{\frac{T}{\mu}}$ Rearranging for tension ($T$): $T = v^2 \mu$

Calculation: Using the values of $v$ and $\mu$: $T = (78.75 \text{ m/s})^2 \times (4.0 \times 10^{-2} \text{ kg/m})$ $T = 6201.5625 \times 4.0 \times 10^{-2} \text{ N}$ $T = 248.0625 \text{ N}$

Final Answer (b): The tension in the string is approximately $248 \text{ N}$.

Given: Frequency of the source, $f = 340 \text{ Hz}$ First resonating length, $L_1 = 25.5 \text{ cm} = 0.255 \text{ m}$ Second resonating length, $L_2 = 79.3 \text{ cm} = 0.793 \text{ m}$

To Find: The speed of sound in air, $v$.

Analysis: The tube with a piston acts as an air column closed at one end (the piston) and open at the other. For such a tube, resonance occurs when the length of the air column is an odd multiple of a quarter wavelength. $L = (2n+1) \frac{\lambda}{4}, \quad \text{where } n = 0, 1, 2, ...$ Let the first resonance at $L_1$ correspond to $n$, and the second resonance at $L_2$ correspond to the next mode, $n+1$. $L_1 = (2n+1) \frac{\lambda}{4}$ $L_2 = (2(n+1)+1) \frac{\lambda}{4} = (2n+3) \frac{\lambda}{4}$

Formula: Subtracting the first equation from the second gives the difference in length between two consecutive resonances: $L_2 - L_1 = (2n+3) \frac{\lambda}{4} - (2n+1) \frac{\lambda}{4}$ $L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2}$ Therefore, the wavelength can be found as $\lambda = 2(L_2 - L_1)$. The speed of sound is then given by $v = f \lambda$.

Calculation: First, calculate the wavelength ($\lambda$): $\lambda = 2(L_2 - L_1) = 2(0.793 \text{ m} - 0.255 \text{ m})$ $\lambda = 2(0.538 \text{ m}) = 1.076 \text{ m}$ Now, calculate the speed of sound ($v$): $v = f \lambda = (340 \text{ Hz}) \times (1.076 \text{ m})$ $v = 365.84 \text{ m/s}$

Final Answer: The estimated speed of sound in air is approximately $366 \text{ m/s}$.

Given: Length of the steel rod, $L = 100 \text{ cm} = 1.0 \text{ m}$ Fundamental frequency of longitudinal vibrations, $f_1 = 2.53 \text{ kHz} = 2.53 \times 10^3 \text{ Hz}$

To Find: The speed of sound in steel, $v$.

Analysis: The rod is clamped at its middle. For longitudinal vibrations, the clamped point must be a displacement node. In the fundamental mode of vibration, the ends of the rod are free to vibrate and will be displacement antinodes. Therefore, the pattern of the standing wave is Antinode-Node-Antinode. The distance between an antinode and the next node is $\lambda/4$. The total length of the rod is the distance between the two antinodes at the ends. $L = \frac{\lambda}{4} + \frac{\lambda}{4} = \frac{\lambda}{2}$

Formula: From the analysis, the wavelength of the fundamental mode is $\lambda = 2L$. The speed of sound is related to frequency and wavelength by $v = f_1 \lambda$.

Calculation: First, calculate the wavelength ($\lambda$): $\lambda = 2L = 2 \times 1.0 \text{ m} = 2.0 \text{ m}$ Now, calculate the speed of sound ($v$): $v = f_1 \lambda = (2.53 \times 10^3 \text{ Hz}) \times (2.0 \text{ m})$ $v = 5.06 \times 10^3 \text{ m/s} = 5060 \text{ m/s}$

Final Answer: The speed of sound in steel is $5060 \text{ m/s}$ (or $5.06 \text{ km/s}$).

Given: Length of the pipe, $L = 20 \text{ cm} = 0.20 \text{ m}$ Frequency of the source, $f_{source} = 430 \text{ Hz}$ Speed of sound in air, $v = 340 \text{ m/s}$

Formula: For a pipe closed at one end, only odd harmonics are present. The frequencies of the normal modes are given by: $f_n = \frac{(2n+1)v}{4L}, \quad \text{where } n = 0, 1, 2, ...$ The fundamental frequency (first harmonic, n=0) is: $f_1 = \frac{v}{4L}$

Calculation: First, calculate the fundamental frequency: $f_1 = \frac{340 \text{ m/s}}{4 \times 0.20 \text{ m}} = \frac{340}{0.8} = 425 \text{ Hz}$ The possible resonant frequencies are odd multiples of the fundamental frequency: $f_1, 3f_1, 5f_1, ...$ $f_1 = 425 \text{ Hz}$ $f_3 = 3 \times 425 = 1275 \text{ Hz}$ ...and so on. The source frequency is $f_{source} = 430 \text{ Hz}$. This is very close to the fundamental frequency of $425 \text{ Hz}$. Therefore, the pipe will be resonantly excited in its fundamental mode (or first harmonic).

Formula: For a pipe open at both ends, all harmonics are present. The frequencies of the normal modes are given by: $f'_n = \frac{nv}{2L}, \quad \text{where } n = 1, 2, 3, ...$ The fundamental frequency is: $f'_1 = \frac{v}{2L}$

Calculation: Calculate the new fundamental frequency: $f'_1 = \frac{340 \text{ m/s}}{2 \times 0.20 \text{ m}} = \frac{340}{0.4} = 850 \text{ Hz}$ The possible resonant frequencies are integer multiples of this frequency: $850 \text{ Hz}, 1700 \text{ Hz}, ...$ The source frequency of $430 \text{ Hz}$ is not an integer multiple of the fundamental frequency ($850 \text{ Hz}$). Therefore, resonance will not be observed.

Final Answer: For the pipe closed at one end, the first harmonic (fundamental mode) is resonantly excited. No, the same source will not be in resonance with the pipe if both ends are open.

Given: Initial beat frequency, $f_{beat,1} = 6 \text{ Hz}$ Final beat frequency, $f_{beat,2} = 3 \text{ Hz}$ Original frequency of string A, $f_A = 324 \text{ Hz}$

To Find: The frequency of string B, $f_B$.

Analysis: The beat frequency is the absolute difference between the two frequencies: $f_{beat} = |f_A - f_B|$.

Step 1: Determine the possible initial frequencies of B. Initially, $|f_A - f_B| = 6 \text{ Hz}$. Since $f_A = 324 \text{ Hz}$, we have two possibilities for $f_B$: Case 1: $f_B = f_A + 6 = 324 + 6 = 330 \text{ Hz}$ Case 2: $f_B = f_A - 6 = 324 - 6 = 318 \text{ Hz}$

Step 2: Analyze the effect of reducing tension in string A. The frequency of a string is related to its tension $T$ by $f \propto \sqrt{T}$. When the tension in string A is reduced, its frequency $f_A$ decreases. Let the new frequency be $f'_A$, where $f'_A < 324 \text{ Hz}$. The new beat frequency is 3 Hz, so $|f'_A - f_B| = 3 \text{ Hz}$.

Conclusion: The second case is the correct one.

Final Answer: The frequency of B is $318 \text{ Hz}$.

(a) In a sound wave, a displacement node is a pressure antinode and vice versa. A sound wave propagates as compressions and rarefactions. In a standing sound wave:

(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes". Bats use a biological sonar system called echolocation. They emit high-frequency ultrasonic pulses (sound waves beyond the range of human hearing). These pulses travel outwards, reflect off objects in their path, and return as echoes. The bat's brain processes these echoes to build a detailed 'sound map' of its surroundings:

(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes. This distinction is due to the timbre or quality of the sound. When an instrument plays a note, it produces a sound wave that consists of a fundamental frequency (which determines the pitch) and a series of overtones or harmonics (which are integer multiples of the fundamental frequency). The number of overtones present and their relative intensities are unique to each instrument. A violin and a sitar producing the same note (same fundamental frequency) will have different combinations and intensities of harmonics. Our ears and brain perceive this difference in the harmonic content as a difference in the quality of the sound, allowing us to distinguish between the two instruments.

(e) The shape of a pulse gets distorted during propagation in a dispersive medium. A pulse is not a simple sinusoidal wave but a superposition of many waves with different frequencies (or wavelengths). A dispersive medium is one in which the speed of a wave depends on its frequency. When a pulse travels through such a medium, each of its component frequencies travels at a slightly different speed. The faster components get ahead of the slower ones. This change in the relative positions and phases of the component waves causes the overall shape of the pulse to change, or get distorted, as it propagates.